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Chapter 5: Problem 81

You are standing on a bathroom scale in an elevator in a tall building. Your mass is \(64 \mathrm{~kg} .\) The elevator starts from rest and travels upward with a speed that varies with time according to \(v(t)=\left(3.0 \mathrm{~m} / \mathrm{s}^{2}\right) t+\left(0.20 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2} .\) When \(t=4.0 \mathrm{~s},\) what is the reading on the bathroom scale?

Short Answer

Expert verified
The reading on the bathroom scale at \(t=4.0 \mathrm{~s}\) is \(94.0 \mathrm{~kg}\).

Step by step solution

01

Convert speed into acceleration

Since the speed of the elevator changes with time, we need to compute its acceleration. Acceleration is the derivative of speed with respect to time. Given \(v(t)=\left(3.0 \mathrm{~m} / \mathrm{s}^{2}\right) t+\left(0.20 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\), differentiate \(v(t)\) with respect to \(t\) to find \(a(t)\). Using basic calculus, the derivative of \(t\) is 1 and that of \(t^{2}\) is \(2t\). Thus, \(a(t)=3.0 \mathrm{~m} / \mathrm{s}^{2}+0.40 \mathrm{~m} / \mathrm{s}^{3}\cdot t\).
02

Calculate acceleration at given time

Insert \(t=4.0 \mathrm{~s}\) into \(a(t)\) to find the acceleration at that moment. So, \(a(4.0)=3.0 \mathrm{~m} / \mathrm{s}^{2}+0.40 \mathrm{~m} / \mathrm{s}^{3}\cdot 4.0 \mathrm{~s}=3.0 \mathrm{~m} / \mathrm{s}^{2}+1.6 \mathrm{~m} / \mathrm{s}^{2}=4.6 \mathrm{~m} / \mathrm{s}^{2}\).
03

Calculate net force acting on you

The net force acting on a person in an elevator is the sum of the forces due to gravity and the elevator's acceleration. The force due to gravity \(F_g=mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity (typically \(9.8 \mathrm{~m} / \mathrm{s}^{2}\)). So, \(F_g=64 \mathrm{~kg} \cdot 9.8 \mathrm{~m} / \mathrm{s}^{2} = 627.2 \mathrm{~N}\). The force due to the elevator's acceleration \(F_e=ma\), where \(a\) is the elevator's acceleration. So, \(F_e=64 \mathrm{~kg} \cdot 4.6 \mathrm{~m} / \mathrm{s}^{2} = 294.4 \mathrm{~N}\). Combining the two forces, we get \(F_{\text{net}}=F_g+F_e=627.2 \mathrm{~N} + 294.4 \mathrm{~N}=921.6 \mathrm{~N}\).
04

Calculate scale reading

The scale measures the net force. However, it's calibrated in units of mass (kilograms or pounds), assuming the acceleration is due to gravity only. So, to convert the force into mass units which the scale can display, divide the net force by the acceleration due to gravity. Therefore, the scale reading is \(F_{\text{net}} / g = 921.6 \mathrm{~N} / 9.8 \mathrm{~m} / \mathrm{s}^{2} = 94.0 \mathrm{~kg}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law of motion is fundamental in understanding forces and motion. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be expressed through the formula,
  • \( F = ma \)
where:
  • \( F \) is the net force exerted on the object,
  • \( m \) is the mass of the object,
  • \( a \) is the acceleration of the object.

In this exercise, the passenger inside the elevator experiences forces due to gravity and the elevator's accelerating motion. The scale reading reflects these forces.
Newton's Second Law provides the framework to calculate the net force, determining how the elevator's motion impacts what the scale displays.
Calculus in Physics
Calculus is an essential tool in physics to find changes with respect to time, like velocity and acceleration. In this problem, speed is given as a function of time. To find how quickly speed changes鈥攊.e., the acceleration鈥攚e differentiate.
To differentiate the velocity function, \(v(t)=3.0t + 0.20t^2\), you find the derivative:
  • The derivative of \(3.0t\) is \(3.0\),
  • The derivative of \(0.20t^2\) is \(0.40t\).

This gives us an acceleration function \(a(t) = 3.0 + 0.40t\).
Calculus allows us to understand how velocity changes over time and quantify how these changes will impact the scale reading inside the elevator.
Net Force Calculation
Net force is the overall force acting on an object when all individual forces are combined. It dictates acceleration according to Newton's Second Law.
In this problem, two forces act on the passenger:
  • The gravitational force \(F_g = mg\), downward,
  • The force due to elevator's acceleration \(F_e = ma\), upward.

To find the net force, add these forces considering direction:
  • \(F_g = 64 \times 9.8 = 627.2 \) N
  • \(F_e = 64 \times 4.6 = 294.4 \) N

The net upward force becomes \(627.2 + 294.4 = 921.6 \) N. This is what the scale responds to, corresponding to a perceived weight on the scale.
Kinematics Equations
Kinematics is the study of motion without considering forces. It uses equations to describe motion parameters like velocity and acceleration.
In the problem, the given equation for velocity shows how speed changes over time:
  • \(v(t) = 3.0t + 0.20t^2\)
From this, we derive acceleration, a crucial kinematic quantity, using calculus.
Kinematics equations form the backbone of understanding how motion progresses. They are vital for predicting future states of moving objects, like predicting the outcome of the elevator's increasing speed and resultant forces. Through this, we can determine what a scale might read at any given moment as it reacts to dynamic forces in the elevator.

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Most popular questions from this chapter

\(\mathrm{A}\) small car with mass \(0.800 \mathrm{~kg}\) travels at constant speed on the inside of a track that is a vertical circle with radius \(5.00 \mathrm{~m}\) (Fig. E5.45). If the normal force exerted by the track on the car when it is at the top of the track (point \(B\) ) is \(6.00 \mathrm{~N}\), what is the normal force on the car when it is at the bottom of the track (point \(A\) )?

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius \(2.5 \mathrm{~m} .\) The cylinder started to rotate, and when it reached a constant rotation rate of \(0.60 \mathrm{rev} / \mathrm{s},\) the floor dropped about \(0.5 \mathrm{~m}\). The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (Note: When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the wall to the floor.

Runway Design. A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is \(700 \mathrm{~kg}\), and the total resistance (air drag plus friction with the runway on each may be assumed constant and equal to \(2500 \mathrm{~N}\). The tension in the towrope between the transport plane and the first glider is not to exceed \(12,000 \mathrm{~N}\). (a) If a speed of \(40 \mathrm{~m} / \mathrm{s}\) is required for takeoff, what minimum length of runway is needed? (b) What is the tension in the towrope between the two gliders while they are accelerating for the takeoff?

The Trendelenburg Position. After emergencies with major blood loss, a patient is placed in the Trendelenburg position, in which the foot of the bed is raised to get maximum blood flow to the brain. If the coefficient of static friction between a typical patient and the bed sheets is \(1.20,\) what is the maximum angle at which the bed can be tilted with respect to the floor before the patient begins to slide?

Genesis Crash. On September \(8,2004,\) the Genesis spacecraft crashed in the Utah desert because its parachute did not open. The \(210 \mathrm{~kg}\) capsule hit the ground at \(311 \mathrm{~km} / \mathrm{h}\) and penetrated the soil to a depth of \(81.0 \mathrm{~cm}\). (a) What was its acceleration (in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's) assumed to be constant, during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) How long did this force last?
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